How to write Euler product and Zeta function for $K = \mathbb{Q}(\sqrt{3})$ For general zeta number fields, I have found a carefully written statement of the Dedekind zeta function and it's Euler product:
$$ \zeta_L(s) = \sum_{\mathfrak{a}\subseteq \mathcal{O}_L} \frac{1}{N_{L/\mathbb{Q}}(\mathfrak{a})^s} = \prod_{\mathfrak{p}\subseteq \mathcal{O}_L} \left( 1 - \frac{1}{N_{L/\mathbb{Q}}(\mathfrak{p})^s}\right)^{-1} $$
If we use a quadratic field, e.g. $L = \mathbb{Q}(\sqrt{3})$ then I believe the ring of integers are to adjoin the square root:
$$ \mathcal{O}_{\mathbb{Q}(\sqrt{3})} = \mathbb{Z}\big[\sqrt{3}\big] = \{ a + b \sqrt{3} : a,b \in \mathbb{Z}\}$$
I would write the zeta function as the sum of the numbers in this set:
$$ \zeta(s) = \sum'_{(a,b) \in \mathbb{Z}^2} \frac{1}{(a^2 - 3b^2)^s} $$
This is not right... Ideals are indexed by these elements except now there is the possibility of having infinitely many numbers of size $1$:
$$ N_{\mathbb{Q}(\sqrt{3})/\mathbb{Q}}\big[(2 - \sqrt{3})^n \big] = (2 - \sqrt{3})^n(2 + \sqrt{3})^n = (2^2 - 3)^n = 1^n = 1$$
So... all the terms will appear infinitely many times.  What is the correct subset: How can I index the ideals of $\mathcal{O}_{\mathbb{Q}(\sqrt{3})}$ as pairs of integers $(a,b) \in \mathbb{Z}^2$?  My best guess is to find a way to index:
$$ \mathbb{Z}[\sqrt{3}]^\times / \{(2 - \sqrt{3})^n : n \in \mathbb{Z}\}  $$
There migth not be a nice way to index that set.  As for the Euler product, I'd like to know which primes split over $\mathbb{Q}(\sqrt{-3})$. 
Quadratic reciprocity (or possibly a geometry of numbers argument) wouls  say:
$$  \big(  \frac{3}{p}\big)  = (-1)^{\frac{p-1}{2}\frac{3-1}{2}}\big( \frac{p}{3}\big) = \left\{
\begin{array}{rl} 
1 &  p \equiv 1 \pmod 3\\
-1 &  p \equiv 0,2 \pmod 3
\end{array}  \right.$$
The closest (certainly wrong) statement of the Euler product I could come up with is:
$$ \sum'_{(a,b) \in \mathbb{Z}^2} \frac{1}{(a^2 - 3b^2)^s} = \prod_{p \equiv 1 \pmod 3} \frac{1}{1 - \frac{1}{p^{2s} }}
\prod_{p \equiv 0,2 \pmod 3} \frac{1}{1 - \frac{1}{p^{2s} }} $$
This is the progress that I have so far, hope I have given the idea of what I am looking for.  Appreciate advice or corrections.
 A: Dedekind zeta functions of non-imaginary quadratic number fields are more complicated. There is how you'll show $$\zeta_{\mathbb{Q}(\sqrt{3})}(s)= \sum_{I \subset \mathbb{Z}[\sqrt{3}]} N(I)^{-s}= \!\!\!\!\!\!\!\!\!\!\!\! \sum_{n+m\sqrt{3} \in \mathbb{Z}[\sqrt{3}]^*/\mathbb{Z}[\sqrt{3}]^\times}\!\!\!\!\!\! \!\!\!  N(n+m\sqrt{3})^{-s} = \!\!\!\! \!\!\!\!\!\!\sum_{n+m\sqrt{3} \in \mathbb{Z}[\sqrt{3}]^*/(2-\sqrt{3})^\mathbb{Z}}\!\!\!\!\!\!  |n^2-3m^2|^{-s} $$ still comes from a theta function, allowing to obtain the analytic continuation and  the functional equation. 
Let for $(x,y) \in (0,\infty)^2$
$$\vartheta(x,y) = \!\!\!\!\!\!\! \sum_{n+m\sqrt{3} \in \mathbb{Z}[\sqrt{3}]^*/(2-\sqrt{3})^\mathbb{Z}}\!\!\!\!\!\! \!\!\! e^{- x |n-m \sqrt{3}|^2-y |n+m \sqrt{3}|^2}$$
$$\Theta(x,y) = \sum_{n+m\sqrt{3} \in \mathbb{Z}[\sqrt{3}]^*} e^{-x |n-m \sqrt{3}|^2-y |n+m \sqrt{3}|^2} = 2\sum_{k=-\infty}^\infty \vartheta(x|2-\sqrt{3}|^{2k},y|2+\sqrt{3}|^{2k})$$
Note $\int_0^\infty x^{s-1} e^{-ax}dx = a^{-s} \Gamma(s)$ means
$$\iint_{(0,\infty)^2} (xy)^{s-1} \vartheta(x,y)dxdy =\Gamma(s)^2  \!\!\!\!\!\!\!\!\!\!\!\! \sum_{n+m\sqrt{3} \in \mathbb{Z}[\sqrt{3}]^*/(2-\sqrt{3})^\mathbb{Z}}\!\!\!\!\!\! \!\!\!\!\!\!\!\!\! |n-m \sqrt{3}|^{-2s}|n+m \sqrt{3}|^{-2s} = \Gamma(s)^2 \zeta_{\mathbb{Q}(\sqrt{3})}(2s) $$
The claim is that there is some simply connected subset $S \subset (0,\infty)\times (0,\infty)$ such that $$(0,\infty)\times (0,\infty) = \bigcup_{k=-\infty}^\infty S \cdot  ( |2-\sqrt{3}|^{2k},|2+\sqrt{3}|^{2k})$$
where $\cdot$ means $(a,b) \cdot (c,d) = (ac,bd)$. 
Proof : take $T$ a strip in the plane being a fundamental domain for $\frac{\mathbb{R}^2 }{ \mathbb{Z}\  (\log |2-\sqrt{3}|^2,\log |2+\sqrt{3}|^2)}$ then $S = \exp(T)$.
One can check that
$$\iint_{(0,\infty)^2} (xy)^{s-1} \vartheta(x,y)dxdy=\frac12\iint_S  (xy)^{s-1} \Theta(x,y)dxdy$$
